`KMnO_4` reacts with `Na_2S_2O_3` in acidic, strongly basic and aqueous (neutral) media. `100mL` of `LMnO_4` reacts with 100 mL of 0.1 M `Na_2S_2O_3` in acidic, basic and neutral media.
Q. The molarity `(M)` of `KMnO_4` solution in the acidic medium is

Options

A. 0.2 M
B. 0.02 M
C. 0.4 M
D. 0.04 M

Explanation

To find the molarity of the `KMnO_4` solution in acidic medium, we can follow these steps:

### Step 1: Understand the Reaction
In acidic medium, `KMnO_4` (potassium permanganate) is reduced from Mn in the +7 oxidation state to Mn in the +4 oxidation state. The half-reaction can be represented as:
\[ \text{Mn}^{7+} + 3e^- \rightarrow \text{Mn}^{4+} \]
This indicates that 1 mole of `KMnO_4` will gain 3 moles of electrons (n-factor = 3).

### Step 2: Calculate Milliequivalents
Milliequivalents (meq) can be calculated using the formula:
\[ \text{Milliequivalents} = \text{Molarity (M)} \times \text{Volume (L)} \times \text{n-factor} \]

For `KMnO_4`:
- Volume = 100 mL = 0.1 L
- n-factor = 3
- Let the molarity of `KMnO_4` be \( M \).

Thus, the milliequivalents of `KMnO_4` is:
\[ \text{Milliequivalents of } KMnO_4 = M \times 0.1 \times 3 \]

For `Na_2S_2O_3` (sodium thiosulfate):
- Volume = 100 mL = 0.1 L
- Molarity = 0.1 M
- n-factor = 8 (as it can donate 8 electrons in total).

Thus, the milliequivalents of `Na_2S_2O_3` is:
\[ \text{Milliequivalents of } Na_2S_2O_3 = 0.1 \times 0.1 \times 8 = 0.08 \]

### Step 3: Set Up the Equation
Since the milliequivalents of `KMnO_4` will equal the milliequivalents of `Na_2S_2O_3`, we can set up the equation:
\[ M \times 0.1 \times 3 = 0.08 \]

### Step 4: Solve for Molarity
Now, we can solve for \( M \):
\[ M \times 0.3 = 0.08 \]
\[ M = \frac{0.08}{0.3} \]
\[ M = 0.2667 \, \text{M} \]

### Step 5: Round the Answer
Rounding to two decimal places, we find:
\[ M \approx 0.27 \, \text{M} \]

### Final Answer
The molarity of the `KMnO_4` solution in acidic medium is approximately **0.27 M**.

---